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Source: — Data Sufficiency |
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by neelgandham » Thu Jan 12, 2012 9:20 am
----P---R----M----S-----
In the number line, R is zero, M is midpoint between P and S, and PR=1/3PS. What is P?
Let P = -b,and S = a then
PR=1/3PS, implies b = (1/3)*(a+b), 2b = a
M = (a-b)/2 as M is midpoint between P and S. So M = a/4
So, P = -a/2, S = a, M = a/4
Now the question can be rephrased to What is the value of -a/2 ?
1). M is 1.5
M = a/4 = 1.5, i.e -a/2(=P) = -3 Sufficient
2). S is 6
S = a = 6, i.e -a/2(=P) = -3 Sufficient

IMO D
Anil Gandham
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by sud21 » Fri Jan 13, 2012 10:21 am
Let PS=x, PM=MS, PR=1/3PS

PM=x/2, MS=x/2

Now m=1.5 so PR=x/3

x/2-x/3 = RM which is 1.5

So P=x/2-x/3 =1.5 , x = 9 PR=x/3 = 1/3*9= -3 because P is left of R which is 0

Since PS =1/3, then RS = 2/3

So, 2/3 of x = 2/3*9=6

S is 6 because RS=2/3

Answer is D.
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